How to make synchronous api calls in node js?

Question

When I run this code:

import fetch from 'node-fetch';

const url = "https://jsonplaceholder.typicode.com/todos/1";
const get = async () => {
    try {
        let response = await fetch(url);
        let res = await response.json();
        console.log(res);
    } catch (err) {
        console.error(err);
    }
};

(async function () {
    await get();
})();

console.log("I am outside");

I get the following output:

$ node index.js
I am outside
{ userId: 1, id: 1, title: 'delectus aut autem', completed: false }

Why I am not getting output in reverse order, even though I have await for async functions?

Answer 1

This is awaited:

await get()

But this is not:

(async function(){
  await get()
})();

If you're using a version of Node which supports top-level await, you can await it:

await (async function(){
  await get()
})();

Or follow up the Promise with a callback:

(async function(){
  await get()
})().then(() => {
  console.log('I am outside');
});

Alternatively, you can move your logic into the IIFE:

(async function(){
  await get();
  console.log('I am outside');
})();

Answer 2

This part is asynchronous

( async function(){
    await get()
})();

I can think of two ways to get the text to display first;

1. Move it before the async-function call
2. move it into the function call;

Answer 3

In my experience, as David said, you should put code that you want to run after asynchronous code in the defined earlier function's scope. Everything synchronous out of scope will run first

Answer 4

Because this function

( async function(){
    await get()
})();

run in asynchronous, so the console.log will print first If you want run the get() first, you should do something like

( async function() {
  await get();
  console.log('I am outside')
})

or another one

( async function() {
  await get();
}).then( () => {
  console.log('I am outside');
})